Calculate ${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}({s}^{2}+{a}^{2})}\right\}$ using the convolution theorem $\mathcal{L}\{f\ast g\}=\mathcal{L}\{f\}\cdot \mathcal{L}\{g\}$

My approach:

$$f(t)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}}\right\}=t$$

$$g(t)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}+{a}^{2}}\right\}=\frac{1}{a}\mathrm{sin}(at)$$

$${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}({s}^{2}+{a}^{2})}\right\}=f\ast g={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f(t-\tau )g(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau =\frac{1}{a}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}(t-\tau )\mathrm{sin}(a\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau $$

but the last integral is clearly divergent.

My approach:

$$f(t)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}}\right\}=t$$

$$g(t)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}+{a}^{2}}\right\}=\frac{1}{a}\mathrm{sin}(at)$$

$${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}({s}^{2}+{a}^{2})}\right\}=f\ast g={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f(t-\tau )g(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau =\frac{1}{a}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}(t-\tau )\mathrm{sin}(a\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau $$

but the last integral is clearly divergent.